Related rates

There are many times when two variable quantities always have a fixed relationship with one another. This is often expressed by an equation that the two quantities must satisfy, regardless of their values. Then if one changes, the other is forced to change in a predictable way, to keep the equation true. These variables will be called related variables.

Examples: 1. Liquid is poured into a cylindrical tank whose base radius is 3 ft′ . Then the height h of liquid in the tank and the volume V of that liquid are always related by the equation

9V hπ=

2. A 10 ft. ladder is placed against a wall. The height y of the top and the distance x of the bottom from the wall are always related by the equation 2 2 100x y+ =

x

y

3. A point on the curve y = x2 is chosen and the x and ycoordinates are recorded. Wherever the point, x and y are always related by that equation.

Now suppose that several related variables are changing with time. The rates of change of the variables must also be related.In related rate problems, information on the values of the related variables, and information on the rate of change of all but one with time are given. The fundamental problem is to find the rate of change of the remaining variable.

This is done by writing down the equation that relates the variables, and differentiating both sides of it with respect to time t, using the chain rule when needed.

Then substitute in known values, and solve for the remaining rate.

Example. A 10 ft. ladder is placed against a wall. If the bottom is 6 ft. from the wall, and is moving away from the wall at the rate of 4 inches per second, how fast is the top of the ladder sliding down the wall?

2 2 100 (1)x y+ =

Solution. As we showed before, the distance x of the base of the ladder from the wall, and the height y of the top of the ladder are always related by the equation

Differentiating both sides with respect to t, we get

2 2 0 (2)dx dyx ydt dt

+ =

The only quantity in equation (2) that is unknown is dydt

We solve (2) for this quantity. We have

2 2dy dxy xdt dt

=− 6 1 1 ft./sec.8 3 4

dy xdxdt y dt

=− =− × =−so

Since x = 6 ft., the equation (1) tells us that y = 8 ft.. We also know that

1 ft/sec.3

dxdt

=

Basic Strategy

Step1. (optional) Draw a figure and label the quantities that vary.

Step2. Find an equation that relates the variable quantities.

Step3. Differentiate the equation found in Step 2 with respect to time, using the chain rule and when needed.

Step4. Use information provided in the problem to find all quantities in the equation of step3 except one rate of change, and then solve for that rate.

We can now identify a basic strategy for solving related rate problems.

Solution. The equation relating radius and area of a circle is2A rπ=

Taking derivatives of both sides with respect to time produces the equation: 2dA drr

dt dtπ=

Thus, at the moment in question, we have

2 (5)(2) 20dAdt

π π= =

cm/sec.

Example. At a certain instant, the radius of a circle is 5 cm and is increasing at the rate of 2 cm/sec. At that instant, whatis the rate of change of the area of the circle.

Example. Let l be the length of the diagonal of a rectangle whose sides have lengths x and y, and assume that x and y vary with time. If x increases at a constant rate of ½ ft/sec, and ydecreases at a constant rate of ¼ ft/sec., how fast is the size of the diagonal changing when x = 3 ft. and y = 4 ft.? It it increasing or decreasing?

Solution.

2 2 2l x y= +

The above diagram shows the situation. The equation relating the variables is:

l

x

y

2 2 2dl dx dyl x ydt dt dt

= +

Taking the derivative with respect to t, we have:

When x = 3 and y = 4, we have 9 16 5l = + =

Thus

3 1 4 15 2 5 4

3 2 110 10 10

dl xdx ydydt l dt l dt

= + = × + ×−

= − = ft/sec.

Since this number is positive, the diagonal is increasing.

Example. Let A be the area of a rectangle whose sides have lengths x and y, and assume that x and y vary with time. If xincreases at a constant rate of ½ ft/sec, and y decreases at a constant rate of ¼ ft/sec., how fast is the area changing when x = 3 ft. and y = 4 ft.? It it increasing or decreasing?

Solution.

The equation relating the variables is: A = xy.

x

yA

dA dx dyy xdt dt dt

= +

Taking the derivative with respect to t, we have:

Thus

1 14 32 4

3 524 4

dA dx dyy xdt dt dt

= + = × − ×

= − = ft/sec.

Since this number is positive, the area is increasing.

Example. A spherical balloon is inflated in such a way that the volume is increasing at a rate of 3 ft3/min. How fast is the diameter increasing when the radius is 1 ft.?

Solution. The volume V of the balloon and it’s diameter a are related by the equation

3 34 4323 3 6a aV r ππ π = = =

Differentiating, we have

2 236 2

dV a da a dadt dt dt

π π= =2

2da dVdt dtaπ

=or

When r = 1, a = 2, so

2 3(3) ft/min4 2

dadt π π

= =

Solution. The situation is shown in the figure below.

2 2 217x y+ =

17 ft y

x

The appropriate equation is

Example. A 17 ft ladder is placed against a wall. If the bottom is pushed along the ground towards the wall at a constant rate of 5 ft./sec., how fast will the top of the ladder be moving up the wall when it is 8 ft. from the ground?

2 2 0dx dyx ydt dt

+ =

17 ft y

x

Taking derivative with respect to time, we get

or dy xdxdt y dt

=−

When y = 8, 2 217 8 225 15.x= − = =

Thus 15 75( 5)8 8

dy xdxdt y dt

=− =− − = ft./sec.

Example. A softball diamond is 60 ft on a side. If the runner is moving toward second base at 25 ft./sec, how fast is her distance from home plate increasing when she is 10 ft from second base?

l x

60

60

Let x be the distance of the runner from first base at a given time, and let l be the corresponding distance from home plate. Clearly 2 2 260x l+ =

l x60

2 0 2dx dlx ldt dt

+ =

Differentiating, we find that

or .dl xdxdt l dt

= When the runner is 10 ft.

from second base, x = 50, and so 2 250 60 6100 10 61.l = + = =

Thus we have 50 125(25) ft./sec.10 61 61

dldt

= =

2 2 260x l+ =

Example. Grain pours from a chute at the rate of 8 cubic ft. per minute. It forms a conical pile for which the height is always twice the radius of the base. Find the rate of increase of the height of the pile when the height is 6ft.

radius r

height hVolume = V.

radius rheight h

Clearly,V and h are related by the formula2 31 12

23 3 12h hV r h h ππ π = = =

Volume = V.

2 2312 4

dV h dh h dhdt dt dt

π π= =

Differentiating, we have

When h = 6, 4 4 4 8(8)

2 36 36 9dh dV dVdt dt dth π π ππ

= = = =

Example. An aircraft is climbing at a 30° angle to the horizontal. How fast is the aircraft gaining altitude if its speed is 500 mi./hr.

We know that

30°altitude h

distance s

sin( )h s θ= and that 500.dsdt

= Thus

1sin( ) 2502

dh ds dsdt dt dt

θ= = = mi./hr.

Example. A particle is moving along the curve whose equation is

Assume that the x-coordinate is increasing at the rate of 6 units/sec. When the particle is at the point (1, 2). At what rate it the y-coordinate changing at that instant? Is it increasing or decreasing?

3 82 51

xy

y=

+

Solution. We differentiate the equation of the curve.

( )( ) ( )( )( )

2 3 3 21 10

221

y xy xy y

y

′ ′+ − +

=

+

( ) ( )( )

2 3 2 31 3 20

221

dx dy dyy y xy xy ydt dt dt

y

+ + − =

+

2 3 2 31 3 2 0dx dy dyy y xy xy ydt dt dt

+ + − =

( )3 2 2 322 1 3 1dy dxdyxy y y xy y ydt dtdt

− + = +

Substituting known values produces

( )( )32 60 5 (6)(8) 240dy dydt dt

− = =

240 6028 7

dydt

=− =−